Number of ways to choose an integer such that there are exactly K elements greater than it in the given array

Given an array arr[] of N elements and an integer K, the task is to find the number of ways of choosing an integer X such that there are exactly K elements in the array that are greater than X.

Examples:

Input: arr[] = {1, 3, 4, 6, 8}, K = 2
Output: 2
X can be chosen as 4 or 5

Input: arr[] = {1, 1, 1}, K = 2
Output: 0
No matter what integer you choose as X, it’ll never have exactly 2 elements greater than it in the given array.

Approach:
We count total number of distinct elements in the array. If the count of distinct elements is less than or equal to k, then 0 permutations possible. Else count is equal to number of distinct elements minus k.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to returns the required count of integers
int countWays(int n, int arr[], int k)
{
  
    if (k <= 0 || k >= n)
        return 0;
  
    unordered_set<int> s(arr, arr+n);
    if (s.size() <= k)
       return 0;
  
    // Return the required count
    return s.size() - k;
}
  
// Driver code
int main()
{
    int arr[] = { 100, 200, 400, 50 };
    int k = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countWays(n, arr, k);
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to returns the 
// required count of integers
static int countWays(int n, int arr[], int k)
{
  
    if (k <= 0 || k >= n)
        return 0;
    Set<Integer> s = new HashSet<Integer>();
    for(int i = 0; i < n; i++)
        s.add(arr[i]);
          
    if (s.size() <= k)
        return 0;
  
    // Return the required count
    return s.size() - k;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 100, 200, 400, 50 };
    int k = 3;
    int n = arr.length;
    System.out.println(countWays(n, arr, k));
}
}
  
// This code id contributed by
// Surendra_Gangwar

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Python3

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# Python 3 implementation of the approach 
  
# Function to returns the required count of integers 
def countWays(n, arr, k) :
  
    if (k <= 0 or k >= n) :
        return 0
  
    s = set()
    for element in arr :
        s.add(element)
          
    if (len(s) <= k) :
        return 0
  
    # Return the required count 
    return len(s) - k; 
  
  
# Driver code 
if __name__ == "__main__" :
      
    arr = [ 100, 200, 400, 50
    k = 3
    n = len(arr) ;
    print(countWays(n, arr, k)) 
  
# This code is contributed by Ryuga

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
  
class GFG
{
  
    // Function to returns the 
    // required count of integers
    static int countWays(int n, int []arr, int k)
    {
  
        if (k <= 0 || k >= n)
            return 0;
        HashSet<int> s = new HashSet<int>();
        for(int i = 0; i < n; i++)
            s.Add(arr[i]);
  
        if (s.Count <= k)
            return 0;
  
        // Return the required count
        return s.Count - k;
    }
  
    // Driver code
    public static void Main(String []args)
    {
        int []arr = { 100, 200, 400, 50 };
        int k = 3;
        int n = arr.Length;
        Console.WriteLine(countWays(n, arr, k));
    }
}
  
// This code is contributed by Rajput-Ji

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PHP

= $n)
return 0;

$s = array();
foreach ($arr as $value)
array_push($s, $value);
$s = array_unique($s);

if (count($s) <= $k) return 0; // Return the required count return count($s) - $k; } // Driver code $arr = array(100, 200, 400, 50); $k = 3; $n = count($arr); print(countWays($n, $arr, $k)); // This code is contributed by mits ?>

Output:

1

Time Complexity: O(N * log(N))



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